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A loudspeaker at the origin emits a 140Hz tone on a day when the speed of sound is 340 m/s. The phase difference between two points on the x-axis is 5.1rad .What is the distance between these two points?

Respuesta :

Samueloladosu37

Answer:1.97m

Explanation:

the wave period

T = 1 / f

T = 1 / 140

T = 0.00714 S

Each wave length is

340 m/s(0.00714)s = 2.4276 m

5.1rad / 2π rad/cycle = 0.8116cycles

d = 0.8116 cycle (2.4276 m/cycle) = 1.970m

okpalawalter8

The distance between these two points is mathematically given as

d = 1.970m

What is the distance between these two points?

Question Parameter(s):

A loudspeaker at the origin emits a 140Hz tone on

a day when the speed of sound is 340 m/s.

Generally, the equation for the wave period   is mathematically given as

T = 1 / f

Therefore

T = 1 / 140

T = 0.00714 S

Hence Wavelenght

wave length is 340 m/s(0.00714)s = 2.4276 m

5.1rad / 2π  = 0.8116cycles

d = 0.8116 cycle

d = 1.970m

In conclusion, the distance is

d = 1.970m

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